# What are the points of inflection, if any, of f(x)=x^4-x^2+5 ?

Dec 24, 2015

Inclection points are $\left(\frac{\sqrt{6}}{6} , \frac{187}{36}\right)$ and $\left(- \frac{\sqrt{6}}{6} , \frac{187}{36}\right)$.

#### Explanation:

$\textcolor{w h i t e}{\times} f \left(x\right) = {x}^{4} - {x}^{2} + 5$

$\implies \frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = 4 {x}^{3} - 2 x$
$\implies {d}^{2} / {\mathrm{dx}}^{2} \left[f \left(x\right)\right] = 12 {x}^{2} - 2$

For ${d}^{2} / {\mathrm{dx}}^{2} \left[f \left(x\right)\right] = 0$,
$\textcolor{w h i t e}{\times} 12 {x}^{2} - 2 = 0$

$\implies x = \pm \frac{\sqrt{6}}{6}$

For $x = \frac{\sqrt{6}}{6}$,
$\textcolor{w h i t e}{\times} y = {\textcolor{red}{\left(\frac{\sqrt{6}}{6}\right)}}^{4} - {\textcolor{red}{\left(\frac{\sqrt{6}}{6}\right)}}^{2} + 5$
$\textcolor{w h i t e}{\times x} = \frac{1}{6} ^ 2 - \frac{1}{6} + 5$
$\textcolor{w h i t e}{\times x} = \frac{187}{36}$

For $x = - \frac{\sqrt{6}}{6}$,
$\textcolor{w h i t e}{\times} y = \frac{187}{36}$

Therefore inclection points are $\left(\frac{\sqrt{6}}{6} , \frac{187}{36}\right)$ and $\left(- \frac{\sqrt{6}}{6} , \frac{187}{36}\right)$. 